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<h1 class="heading"><a href="MATH-2023-OPDE.html"><span class="title">MATH 2023: Ordinary and Partial Differential Equations</span></a></h1>
<p class="byline">Xiaoyi Chen and Wei Zhang</p>
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<li class="link frontmatter"><a href="meta_frontmatter.html" data-scroll="meta_frontmatter" class="internal"><span class="title">Front Matter</span></a></li>
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<a href="ch_first.html" data-scroll="ch_first" class="internal"><span class="codenumber">1</span> <span class="title">Introduction</span></a><ul>
<li><a href="sec_1-intro.html" data-scroll="sec_1-intro" class="internal">Classification of Differential Equations</a></li>
<li><a href="sec_2-intro.html" data-scroll="sec_2-intro" class="internal">Linear and Nonlinear Equation</a></li>
<li><a href="sec_3-intro.html" data-scroll="sec_3-intro" class="internal">Geometrical Aspect</a></li>
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<a href="ch_second.html" data-scroll="ch_second" class="internal"><span class="codenumber">2</span> <span class="title">First Order Ordinary Differential Equations</span></a><ul>
<li><a href="sec2_1.html" data-scroll="sec2_1" class="active">Linear Equations</a></li>
<li><a href="sec2_2.html" data-scroll="sec2_2" class="internal">Further Discussion of Linear Equations (For reading only)</a></li>
<li><a href="sec2_3.html" data-scroll="sec2_3" class="internal">Separable Equations</a></li>
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<li><a href="sec2_5.html" data-scroll="sec2_5" class="internal">Applications of modeling with first order ODE(For reading only)</a></li>
<li><a href="sec2_6.html" data-scroll="sec2_6" class="internal">Exact Equations and Integrating Factors</a></li>
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<a href="ch_third.html" data-scroll="ch_third" class="internal"><span class="codenumber">3</span> <span class="title">third Order Linear Equations</span></a><ul>
<li><a href="sec3_1.html" data-scroll="sec3_1" class="internal">Homogeneous equations with constant coefficient</a></li>
<li><a href="sec3_2.html" data-scroll="sec3_2" class="internal">Fundamental Solutions of Linear Homogeneous Equations</a></li>
<li><a href="sec3_3.html" data-scroll="sec3_3" class="internal">Linear Independence and Wronskian</a></li>
<li><a href="sec3_4.html" data-scroll="sec3_4" class="internal">Complex roots of the characteristic equations</a></li>
<li><a href="sec3_5.html" data-scroll="sec3_5" class="internal">Repeated Roots: Reduction of Order</a></li>
<li><a href="sec3_6.html" data-scroll="sec3_6" class="internal">Non-homogeneous Equations and Method of Undetermined Coefficients</a></li>
<li><a href="sec3_7.html" data-scroll="sec3_7" class="internal">Variation of Parameters</a></li>
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<li><a href="sec4_1.html" data-scroll="sec4_1" class="internal">General Theory of the <span class="process-math">\(n\)</span>-th Order Linear Equations</a></li>
<li><a href="sec4_2.html" data-scroll="sec4_2" class="internal">Homogeneous Equations with Constant Coefficients</a></li>
<li><a href="sec4_3.html" data-scroll="sec4_3" class="internal">The Method of Undetermined Coefficients</a></li>
<li><a href="sec4_4.html" data-scroll="sec4_4" class="internal">The Method of Variation of Parameters</a></li>
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<a href="ch_five.html" data-scroll="ch_five" class="internal"><span class="codenumber">5</span> <span class="title">Series Solutions of Second Order Linear Equations</span></a><ul>
<li><a href="sec5_1.html" data-scroll="sec5_1" class="internal">Brief Review on Power Series</a></li>
<li><a href="sec5_2.html" data-scroll="sec5_2" class="internal">Introduction</a></li>
<li><a href="sec5_3.html" data-scroll="sec5_3" class="internal">Series Solutions Near an Ordinary Point</a></li>
<li><a href="sec5_4.html" data-scroll="sec5_4" class="internal">Euler’s Equation</a></li>
<li><a href="sec5_5.html" data-scroll="sec5_5" class="internal">Series Solution near a Regular Singular Point</a></li>
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<a href="ch_six.html" data-scroll="ch_six" class="internal"><span class="codenumber">6</span> <span class="title">System of First Order Linear Equations</span></a><ul>
<li><a href="sec6_1.html" data-scroll="sec6_1" class="internal">Introduction <span class="process-math">\(\&amp;\)</span> Basic Theory</a></li>
<li><a href="sec6_2.html" data-scroll="sec6_2" class="internal">Homogeneous System with Constant Coefficients</a></li>
<li><a href="sec6_3.html" data-scroll="sec6_3" class="internal">Complex Eigenvalues</a></li>
<li><a href="sec6_4.html" data-scroll="sec6_4" class="internal">Repeated Eigenvalues</a></li>
<li><a href="sec6_5.html" data-scroll="sec6_5" class="internal">Fundamental Matrices</a></li>
<li><a href="sec6_6.html" data-scroll="sec6_6" class="internal">Non-homogeneous linear systems</a></li>
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<a href="ch_seven.html" data-scroll="ch_seven" class="internal"><span class="codenumber">7</span> <span class="title">Partial Differential Equations</span></a><ul>
<li><a href="sec7_1.html" data-scroll="sec7_1" class="internal">Two-Point Boundary Value Problems</a></li>
<li><a href="sec7_2.html" data-scroll="sec7_2" class="internal">Eigenvalue Problems</a></li>
<li><a href="sec7_3.html" data-scroll="sec7_3" class="internal">Fourier Series</a></li>
<li><a href="sec7_4.html" data-scroll="sec7_4" class="internal">The Fourier Convergence Theorem</a></li>
<li><a href="sec7_5.html" data-scroll="sec7_5" class="internal">Even and Odd Functions</a></li>
<li><a href="sec7_6.html" data-scroll="sec7_6" class="internal">Introduction to Partial Differential Equations</a></li>
<li><a href="sec7_7.html" data-scroll="sec7_7" class="internal">1D Heat Equation; Solutions by Separation of Variable and Fourier Series</a></li>
<li><a href="sec7_8.html" data-scroll="sec7_8" class="internal">Other Heat Conduction Problems</a></li>
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<a href="ch_eight.html" data-scroll="ch_eight" class="internal"><span class="codenumber">8</span> <span class="title">Laplace transform</span></a><ul>
<li><a href="sec8_1.html" data-scroll="sec8_1" class="internal">What are Laplace Transforms, and Why?</a></li>
<li><a href="sec8_2.html" data-scroll="sec8_2" class="internal">Finding Laplace Transforms</a></li>
<li><a href="sec8_3.html" data-scroll="sec8_3" class="internal">Finding inverse transforms using partial fractions</a></li>
<li><a href="sec8_4.html" data-scroll="sec8_4" class="internal">Solving ODEs and ODE Systems</a></li>
<li><a href="sec8_5.html" data-scroll="sec8_5" class="internal">Step input and Impulse problems</a></li>
<li><a href="sec8_6.html" data-scroll="sec8_6" class="internal">Laplace transform for PDE (heat equation)</a></li>
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<a href="ch_features.html" data-scroll="ch_features" class="internal"><span class="codenumber">9</span> <span class="title">Examples of PreTeXt features</span></a><ul><li><a href="sec_features-blocks.html" data-scroll="sec_features-blocks" class="internal">Environments and Blocks</a></li></ul>
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<li class="link"><a href="solutions-1.html" data-scroll="solutions-1" class="internal"><span class="codenumber">A</span> <span class="title">Selected Hints</span></a></li>
<li class="link"><a href="solutions-2.html" data-scroll="solutions-2" class="internal"><span class="codenumber">B</span> <span class="title">Selected Solutions</span></a></li>
<li class="link"><a href="appendix-1.html" data-scroll="appendix-1" class="internal"><span class="codenumber">C</span> <span class="title">List of Symbols</span></a></li>
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<main class="main"><div id="content" class="pretext-content"><section class="section" id="sec2_1"><h2 class="heading hide-type">
<span class="type">Section</span> <span class="codenumber">2.1</span> <span class="title">Linear Equations</span>
</h2>
<p id="p-16">For a linear first order ODE, <span class="process-math">\(f(x, y)\)</span> need be a linear function of <span class="process-math">\(y\text{,}\)</span> say</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_2.html" id="eq2_1">
\begin{equation}
f(x, y)=-p(x)y+q(x).\tag{2.1.1}
\end{equation}
</div>
<p class="continuation">Then, one has</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_2.html" id="eq2_2">
\begin{equation}
\frac{\textrm{d} y}{\textrm{d} x}+p(x) y=q(x).\tag{2.1.2}
\end{equation}
</div>
<p class="continuation">Equation (<a href="" class="xref" data-knowl="./knowl/eq2_2.html" title="Equation 2.1.2">(2.1.2)</a>) is the first order linear ODE.</p>
<p id="p-17">Consider a particular example:</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_3.html ./knowl/eq2_3.html ./knowl/eq2_3.html ./knowl/eq2_2.html" id="eq2_3">
\begin{equation}
\frac{\textrm{d} y}{\textrm{d} x}+\frac{1}{2} y=\frac{3}{2}.\tag{2.1.3}
\end{equation}
</div>
<p class="continuation">We want to find the solution to <a href="" class="xref" data-knowl="./knowl/eq2_3.html" title="Equation 2.1.3">(2.1.3)</a>. One method is like this:</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_3.html ./knowl/eq2_3.html ./knowl/eq2_3.html ./knowl/eq2_2.html" id="eq2_4">
\begin{equation}
\begin{aligned}
&amp;\frac{\textrm{d} y}{\textrm{d} x}=-\frac{y-3}{2} \rightarrow \frac{1}{y-3} \frac{\textrm{d} y}{\textrm{d} x}=-\frac{1}{2}\rightarrow  \frac{\textrm{d}}{\textrm{d} x} \ln |y-3|=-\frac{1}{2},\\
&amp;\rightarrow \ln |y-3|=-\frac{1}{2} x+C_1 \rightarrow |y-3|=e^{C_1} e^{-\frac{1}{2}x}\\
&amp;\rightarrow y-3=\uuline{\pm e^{C_1}}~ e^{-\frac{1}{2}x} \rightarrow y=3+C e^{-\frac{1}{2}x},
\end{aligned}\tag{2.1.4}
\end{equation}
</div>
<p class="continuation">where <span class="process-math">\(C_1\)</span> is arbitrary constant and <span class="process-math">\(C\)</span> is arbitrary constant not equal to zero. The other method to derive the solution to (<a href="" class="xref" data-knowl="./knowl/eq2_3.html" title="Equation 2.1.3">(2.1.3)</a>) is as follows. Multiplying <span class="process-math">\(e^{\frac{1}{2} x}\)</span> on both sides of (<a href="" class="xref" data-knowl="./knowl/eq2_3.html" title="Equation 2.1.3">(2.1.3)</a>), one has:</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_3.html ./knowl/eq2_3.html ./knowl/eq2_3.html ./knowl/eq2_2.html" id="eq2_5">
\begin{equation}
\begin{aligned}
&amp; \uuline{e^{\frac{1}{2} x} y^{\prime}+\frac{1}{2} e^{\frac{1}{2} x} y}=\frac{3}{2} e^{\frac{1}{2} x} \rightarrow \frac{\textrm{d}}{\textrm{d} x}\left(e^{\frac{1}{2} x} y \right)=\frac{3}{2} e^{\frac{1}{2} x}\\
&amp; \rightarrow e^{\frac{1}{2} x} y =3 e^{\frac{1}{2} x} +C \rightarrow y=3+Ce^{-\frac{1}{2}x},
\end{aligned}\tag{2.1.5}
\end{equation}
</div>
<p class="continuation">where <span class="process-math">\(C\)</span> is an arbitrary constant and <span class="process-math">\(e^{\frac{1}{2} x}\)</span> is called the <dfn class="terminology">integrating factor</dfn>. Through the integrating factor, the resulting equation is readily integrable. For a general linear first order ODE, it can not be solved by the direct method and one may consider using the integrating factor. The difficulty lies in <dfn class="terminology">how to find the integration factor for general case (<a href="" class="xref" data-knowl="./knowl/eq2_2.html" title="Equation 2.1.2">(2.1.2)</a>)?</dfn></p>
<p id="p-18">Consider the general case</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_10.html" id="eq2_6">
\begin{equation}
y^{\prime}+p(x) y=q(x).\tag{2.1.6}
\end{equation}
</div>
<p class="continuation">We multiply it by the integrating factor <span class="process-math">\(u(x)\)</span> (yet to be determined) and have</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_10.html" id="eq2_7">
\begin{equation}
u(x) y^{\prime}+u(x) p(x) y=u(x) q(x).\tag{2.1.7}
\end{equation}
</div>
<p class="continuation">We want <span class="process-math">\(u(x)\)</span> to be such a function that</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_10.html" id="eq2_8">
\begin{equation}
\begin{aligned}
&amp;u(x) y^{\prime}+u(x) p(x) y=\frac{\textrm{d}}{\textrm{d} x} (u(x) y)=u(x) y^{\prime}+\frac{\textrm{d} u(x)}{\textrm{d} x} y,\\
&amp;\rightarrow \frac{1}{u}\frac{\textrm{d} u}{\textrm{d} x}=p(x) \rightarrow \frac{\textrm{d}}{\textrm{d} x} \ln |u(x)|=p(x) \rightarrow \ln |u(x)|=\int p(x) \textrm{d} x+k,
\end{aligned}\tag{2.1.8}
\end{equation}
</div>
<p class="continuation">where <span class="process-math">\(k\)</span> is a constant. The simplest choice is to set <span class="process-math">\(k=0\text{.}\)</span> Then</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_10.html" id="eq2_9">
\begin{equation}
|u(x)|=e^{\int p(x) \textrm{d} x}.\tag{2.1.9}
\end{equation}
</div>
<p class="continuation">Taking <span class="process-math">\(u(x)\)</span> to be non-negative, we have</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_10.html" id="eq2_10">
\begin{equation}
u(x)=e^{\int p(x) \textrm{d} x}.\tag{2.1.10}
\end{equation}
</div>
<p class="continuation">Equation <a href="" class="xref" data-knowl="./knowl/eq2_10.html" title="Equation 2.1.10">(2.1.10)</a> gives one form of integrating factor.</p>
<p id="p-19">With the above integrating factor, (<a href="" class="xref" data-knowl="./knowl/eq2_7.html" title="Equation 2.1.7">(2.1.7)</a>) becomes</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_7.html ./knowl/eq2_10.html ./knowl/eq2_11.html ./knowl/eq2_6.html ./knowl/eq2_11.html ./knowl/eq2_12.html ./knowl/eq2_6.html ./knowl/eq2_11.html ./knowl/eq2_11.html ./knowl/eq2_13.html" id="eq2_11">
\begin{equation}
\begin{aligned}
&amp;\frac{\textrm{d}}{\textrm{d} x} (u(x) y)=u(x) q(x) \rightarrow u(x) y=\int u(x) q(x) \textrm{d} x+C,\\
&amp;\rightarrow y=\frac{\int u(x) q(x) \textrm{d} x+C}{u(x)},
\end{aligned}\tag{2.1.11}
\end{equation}
</div>
<p class="continuation">with <span class="process-math">\(u(x)\)</span> given by (<a href="" class="xref" data-knowl="./knowl/eq2_10.html" title="Equation 2.1.10">(2.1.10)</a>). Equation (<a href="" class="xref" data-knowl="./knowl/eq2_11.html" title="Equation 2.1.11">(2.1.11)</a>) involves an arbitrary constant <span class="process-math">\(C\)</span> and includes every solution to (<a href="" class="xref" data-knowl="./knowl/eq2_6.html" title="Equation 2.1.6">(2.1.6)</a>) and it is called the <dfn class="terminology">general solution</dfn>. Geometrically, (<a href="" class="xref" data-knowl="./knowl/eq2_11.html" title="Equation 2.1.11">(2.1.11)</a>) represents a family of curves, called integral curves. On the other hand, if we impose that</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_7.html ./knowl/eq2_10.html ./knowl/eq2_11.html ./knowl/eq2_6.html ./knowl/eq2_11.html ./knowl/eq2_12.html ./knowl/eq2_6.html ./knowl/eq2_11.html ./knowl/eq2_11.html ./knowl/eq2_13.html" id="eq2_12">
\begin{equation}
y(x_0)=y_0,\tag{2.1.12}
\end{equation}
</div>
<p class="continuation">which is called an initial condition. Then under (<a href="" class="xref" data-knowl="./knowl/eq2_12.html" title="Equation 2.1.12">(2.1.12)</a>), one has</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_7.html ./knowl/eq2_10.html ./knowl/eq2_11.html ./knowl/eq2_6.html ./knowl/eq2_11.html ./knowl/eq2_12.html ./knowl/eq2_6.html ./knowl/eq2_11.html ./knowl/eq2_11.html ./knowl/eq2_13.html" id="eq2_13">
\begin{equation}
y_0=\frac{\int u(x) q(x) \textrm{d} x\Big|_{x=x_0}+C}{u(x_0)},\tag{2.1.13}
\end{equation}
</div>
<p class="continuation">which determines <span class="process-math">\(C\text{.}\)</span> <dfn class="terminology">ODE (<a href="" class="xref" data-knowl="./knowl/eq2_6.html" title="Equation 2.1.6">(2.1.6)</a>) together with initial condition (<a href="" class="xref" data-knowl="./knowl/eq2_11.html" title="Equation 2.1.11">(2.1.11)</a>) is called an initial value problem. The solution (<a href="" class="xref" data-knowl="./knowl/eq2_11.html" title="Equation 2.1.11">(2.1.11)</a>) with <span class="process-math">\(C\)</span> determined by (<a href="" class="xref" data-knowl="./knowl/eq2_13.html" title="Equation 2.1.13">(2.1.13)</a>) is called the particular solution.</dfn></p>
<p id="p-20"><article class="example example-like" id="ex2_1"><a href="" data-knowl="" class="id-ref example-knowl original" data-refid="hk-ex2_1"><h3 class="heading">
<span class="type">Example</span><span class="space"> </span><span class="codenumber">2.1.1</span><span class="period">.</span>
</h3></a></article><div class="hidden-content tex2jax_ignore" id="hk-ex2_1"><article class="example example-like"><dfn class="terminology">Example 1</dfn> Find the general solution of the following ODE:<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y^{\prime}+\frac{y}{x}=3 \cos (2x), \quad x &gt; 0.
\end{equation*}
</div>
<dfn class="terminology">Solution:</dfn> Since <span class="process-math">\(p(x)=\frac{1}{x}\text{,}\)</span> the integrating factor is<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
u(x)=\exp \left(\int p(x) \textrm{d} x\right)=\exp  \left( \int \frac{1}{x} \textrm{d} x \right)=\exp\left(\ln x \right)=x.
\end{equation*}
</div>Then multiplying the integrating factor <span class="process-math">\(u(x)\)</span> on both sides , one has<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
x y^{\prime}+y=3 x \cos (2x)  \rightarrow \frac{\textrm{d}}{\textrm{d} x} (xy)=3 x \cos (2x)  \rightarrow   xy=\int (3 x \cos (2 x)) \textrm{d} x+C=\frac{3}{2} x \sin (2 x)+\frac{3}{4} \cos (2 x)+C.
\end{equation*}
</div>Therefore, the general solution is<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y=\frac{\frac{3}{2} x \sin (2 x)+\frac{3}{4} \cos (2 x)+C}{x}.
\end{equation*}
</div></article></div></p>
<article class="example example-like" id="ex2_2"><a href="" data-knowl="" class="id-ref example-knowl original has-image" data-refid="hk-ex2_2"><h3 class="heading">
<span class="type">Example</span><span class="space"> </span><span class="codenumber">2.1.2</span><span class="period">.</span>
</h3></a></article><div class="hidden-content tex2jax_ignore" id="hk-ex2_2"><article class="example example-like"><p id="p-21"><dfn class="terminology">Example 2</dfn> Find the solution of the following initial value problem:</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_9.html ./knowl/eq2_9.html ./knowl/GPsolution.html" id="eq2_9">
\begin{equation}
x^3 y^{\prime}+4 x^2 y=e^{-x}, \quad y(-1)=0.\tag{2.1.14}
\end{equation}
</div>
<p class="continuation"><dfn class="terminology">Solution</dfn>: The equation is first transformed to</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_9.html ./knowl/eq2_9.html ./knowl/GPsolution.html">
\begin{equation*}
y^{\prime}+\frac{4}{x}y=\frac{e^{-x}}{x^3}.
\end{equation*}
</div>
<p class="continuation">Then one can see <span class="process-math">\(p(x)=\frac{4}{x}\)</span> and the integrating factor is</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_9.html ./knowl/eq2_9.html ./knowl/GPsolution.html">
\begin{equation*}
u(x)=\exp (\int p(x) \textrm{d} x)=\exp (\int \frac{4}{x} \textrm{d} x)=\exp (4 \ln x)=x^4.
\end{equation*}
</div>
<p class="continuation">Multiplying (<a href="" class="xref" data-knowl="./knowl/eq2_9.html" title="Equation 2.1.9">(2.1.9)</a>) with the integrating factor, one has</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_9.html ./knowl/eq2_9.html ./knowl/GPsolution.html">
\begin{equation*}
\begin{aligned}
&amp;x^4 y^{\prime}+4 x^3 y=x e^{-x} \rightarrow \frac{\textrm{d}}{\textrm{d} x} (x^4 y)=x e^{-x} \rightarrow x^4 y=-x e^{-x}-e^{-x}+C.
\end{aligned}
\end{equation*}
</div>
<p class="continuation">Therefore, the general solution is</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_9.html ./knowl/eq2_9.html ./knowl/GPsolution.html">
\begin{equation*}
y=\frac{-x e^{-x}-e^{-x}+C}{x^4}.
\end{equation*}
</div>
<p class="continuation">Applying the initial condition <span class="process-math">\(y(-1)=0\text{,}\)</span> one has <span class="process-math">\(C=0\text{.}\)</span> Therefore, the solution to the initial value problem (<a href="" class="xref" data-knowl="./knowl/eq2_9.html" title="Equation 2.1.9">(2.1.9)</a>) is</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_9.html ./knowl/eq2_9.html ./knowl/GPsolution.html">
\begin{equation*}
y=\frac{-x e^{-x}-e^{-x}}{x^4}.
\end{equation*}
</div>
<p class="continuation"><a href="" class="xref" data-knowl="./knowl/GPsolution.html" title="Figure 2.1.3">Figure 2.1.3</a> shows the general solution and the particular solution.</p>
<figure class="figure figure-like" id="GPsolution"><div class="image-box" style="width: 100%; margin-left: 0%; margin-right: 0%;"><img src="external/./img/GPsolution.eps" class="contained" alt="The general solution and particular solution to Example 2."></div>
<figcaption><span class="type">Figure</span><span class="space"> </span><span class="codenumber">2.1.3<span class="period">.</span></span><span class="space"> </span>The general solution and particular solution to Example 2.</figcaption></figure></article></div></section></div></main>
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